24 : 1/ The Molecular Basis of Nerve Conduction 439 



the same on both sides of the membrane, and G C1 - must be also. Using 

 the subscript i for inside and o for outside, one may write this as 



G Na + . i = £ Na+io 



G C \~ „ = Gcr , 



Referring again to Chapter 21, one can show that in the presence of 

 an electrical potential V, the partial molal Gibbs' free energy of an ionic 

 species is given by 



G = G° + RT\nc + zFV (2) 



where z is the valence of the ion and F is the Faraday. 1 In this expres- 

 sion, G° is the value of G in the standard state with V = 0. For Na + , 

 2 is + 1 and for CI - , z is — 1. Substituting Equation 2 into Equation 1 

 and rearranging, one obtains 



flrin f^j' = zFAF 

 [Na + ] 



(3) 

 tfrinj^LJi = -zFAV 



where AFis the potential difference across the membrane, the outside 

 being positive for AF > 0. 



Adding together the equations in (3) and rearranging shows that 



[Na+MCl-L = [Na + ] [Cl-] (4) 



Because both sides of the membrane have approximately no net charge, 

 it is clear that 



[Na + ] = [Cl-] 



[Na+L = [CI"], + [/»"] 



where [P~] is the proteinate concentration. Because 



[Na + ] 2 > [C1-], whereas [Na + ] = [Cl"] 



Equation 4 allows one to conclude that 



[Na + ] f > [Na + ] and [CI"], < [C\-] 



Accordingly, Equation 3 indicates that AFis positive; that is, the outside 

 of the membrane is positively charged. 



This is the direction of the potential difference observed with fibers 

 such as nerve and muscle. As predicted, the K + ions are at a higher 

 concentration inside than outside the membrane. But the Na + ions 



1 In this chapter, V is used for electrical potential. The same symbol was used 

 for volume in Chapter 21, in discussions of Gibbs' free energy. 



